Problem: A curve is defined by the parametric equations $x=\tan(t)$ and $y=\sin(t)+\dfrac\pi2$. What is $\dfrac{d^2y}{dx^2}$ in terms of $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\sin(t)\tan^3(t)$ (Choice B) B $\cos^5(t)$ (Choice C) C $-3\cos^4(t)\sin(t)$ (Choice D) D $\cos^3(t)\sin(t)$
Explanation: We are asked to find the second derivative of a parametric function. Recall that the first derivative of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ is found with the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ Then, the second derivative is found with this following rule: $\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{\dfrac{d}{dt}\left(\dfrac{v'(t)}{u'(t)}\right)}{u'(t)}$ Let's start by finding $\dfrac{dy}{dx}$. $\dfrac{dy}{dx}=\cos^3(t)$ Now we can find $\dfrac{d^2y}{dx^2}$. $\begin{aligned} \dfrac{d^2y}{dx^2}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dx}{dt}\right)} \\\\ &=\dfrac{\dfrac{d}{dt}\left[\cos^3(t)\right]}{\dfrac{d}{dt}[\tan(t)]} \\\\ &=\dfrac{\left[3\cos^2(t)\right]\cdot\dfrac{d}{dt}[\cos(t)]}{\sec^2(x)} \\\\ &=\dfrac{\left[3\cos^2(t)\right]\cdot[-\sin(t)]}{\sec^2(x)} \\\\ &=\dfrac{-3\cos^2(t)\sin(t)}{\left(\dfrac{1}{\cos^2(x)}\right)} \\\\ &=-3\cos^4(t)\sin(t) \end{aligned}$ In conclusion, $\dfrac{d^2y}{dx^2}=-3\cos^4(t)\sin(t)$.